Compressed Spring Problem

1. Dec 11, 2004

BlackMamba

I have a problem that I just can't seem to figure out.

Here's the problem: The drawing (Click Here) shows a block (m = 1.6 kg) and a spring (k = 325 N/m) on a frictionless incline. The spring is compressed by x_o = 0.35 m relative to it's unstrained position at x = 0 m and then released. What is the speed of the block when the spring is still compressed by x_f = 0.14 m?

I applied the whole Ef = Eo equation but did not get the right answer.

I used sin 30 to find h_f.

So this is what I used:
V_i = 0
V_f = what I'm looking for
h_i = 0
h_f = .5
k = 325
x_i = 0.35
x_f = 0.14

Any help on this problem would be greatly appreciated.

2. Dec 11, 2004

Pyrrhus

This is a problem of application of conservation of mechanical energy

$$\Delta \Omega + \Delta K = 0$$

The energy involved here are kinetic energy, spring potential energy and gravitational potential energy.

At the beginning the spring has spring potential energy and gravitational potential energy, when it reaches the point on the diagram it has spring potential energy and kinetic energy.

3. Dec 11, 2004

BlackMamba

Alrighty.

I know that:
KE = 1/2 mv^2
Spring PE = 1/2 kx
Gravitational PE = mgh

So my equation would look like: 1/2 kx_i + mgh = 1/2 kx_f + 1/2 mv^2

Am I calculating h correctly by using sin 30?

Thanks

4. Dec 11, 2004

Pyrrhus

If you take the 0.14 as your referend for the gravitational potential enegry you should have:

$$info$$
$$x=0.35$$
$$d=0.14$$
$$\theta = 30^{o}$$

$$\frac{1}{2}kx^2 + mg(x-d) \sin \theta = \frac{1}{2}kd^2 + \frac{1}{2}mv^2$$

Last edited: Dec 12, 2004
5. Dec 12, 2004

BlackMamba

Ok well at least I was close. LOL but I don't understand what d is?

6. Dec 13, 2004

BlackMamba

Well, I tried the equation that you gave with those values you specified and I still am not getting the right answer. There must be something I am missing.

The answer I got was 2.33 m/s and it is wrong.

Thanks for helping.

7. Dec 13, 2004

Pyrrhus

I am not getting your numeric value, i get 4.79 m/s